Differentiability — When the Derivative Exists

• If $f$ is differentiable at $x=c$, then $f$ is continuous at $c$.
• The converse is false: $|x|$ is continuous at $0$ but not differentiable there.
• Contrapositive (useful): if $f$ is discontinuous at $c$, it cannot be differentiable at $c$.

$f$ is differentiable at $x=c$ iff the one-sided derivatives match:

• LHD $= \lim_{h\to 0^-}\dfrac{f(c+h)-f(c)}{h}$
• RHD $= \lim_{h\to 0^+}\dfrac{f(c+h)-f(c)}{h}$

Differentiable at $c$ $\iff$ LHD $=$ RHD (and both finite). In practice: differentiate each piece, then equate the two pieces' derivatives at the join (after first checking continuity there). When the one-sided derivatives are INFINITE the function is still not differentiable there, but the graph has a vertical tangent rather than a corner: $y=x^{1/3}$ at $x=0$ has $\dfrac{f(h)-f(0)}{h}=h^{-2/3}\to+\infty$ from both sides — continuous, vertical tangent $x=0$, no derivative. $y=x^{2/3}$ (a cusp: $-\infty$ from the left, $+\infty$ from the right) is the same verdict. "Infinite derivative" in an option list means exactly this case and is a legitimate description, not an error.

Split $|g(x)|$ at the points where $g(x)=0$:

• $|x|$ is continuous but not differentiable at $0$ (LHD $=-1$, RHD $=+1$).
• $|x-a|$ has its corner at $x=a$.
• $x|x|$ equals $-x^2$ for $x<0$ and $x^2$ for $x\ge 0$; both have derivative $0$ at the join, so it is differentiable at $0$ (the trap).

• Between consecutive integers, $\lfloor x\rfloor$ is constant $\Rightarrow$ derivative $=0$ there.
• At every integer it jumps (discontinuous) $\Rightarrow$ not differentiable at integers.
• A function built from $\lfloor x\rfloor$ inherits these jumps unless they cancel; check each integer in the domain.

For a piecewise $f$ differentiable at $x=c$: 1. Continuity: set the two pieces' values equal at $c$. 2. Differentiability: set the two pieces' derivatives equal at $c$. Two equations in the unknown constants — solve simultaneously.

At a point where the usual rules are unsafe, compute $f'(c)=\lim_{h\to 0}\dfrac{f(c+h)-f(c)}{h}$ directly. If the limit exists (and is the same from both sides), that value is the derivative; if it doesn't, $f$ is not differentiable at $c$. The two standard limits these computations need: $\lim_{h\to0}\dfrac{e^h-1}{h}=1$ (so for $f(x)=e^{|x|}$ the RHD at 0 is $+1$ and the LHD is $-1$ — not differentiable), and the small-$u$ approximation $e^u\approx1+u$, which gives $1-e^{-x^2}\approx x^2$ and hence $\sqrt{1-e^{-x^2}}\approx|x|$ — a corner at 0 disguised as a smooth expression. Sibling facts: $\lim_{h\to0}h\ln|h|=0$, $\lim_{h\to0}\dfrac{\sin h}{h}=1$.